Tổng tất cả các giá trị m để đồ thị hàm số y=(x+1)\x^2-2x+m có đúng 1 tiệm cận đứng

$\begin{array}{l} y = \frac{{x + 1}}{{{x^2} - 2x + m}}\\ DK:\,\,\,{x^2} - 2x + m \ne 0\\ Do\,\,thi\,\,ham\,\,so\,\,co\,\,dung\,\,1\,\,tiem\,\,can\,\,dung\\ \Leftrightarrow pt\,\,g\left( x \right) = {x^2} - 2x + m = 0\,\,\,\,co\,\,\,nghiem\,\,\,kep\,\,x \ne - 1\,\,\,\,hoac\,\,\,\,co\,\,\,hai\,\,nghiem\,\,pb\,\,\,trong\,\,do\,\,co\,\,\,1\,\,nghiem\,\,\,x = - 1\\ TH1:\,\,\,pt\,\,{x^2} - 2x + m = 0\,\,\,\,co\,\,\,nghiem\,\,\,kep\,\,x \ne - 1\,\,\,\\ \Leftrightarrow \left\{ \begin{array}{l} \Delta ' = 0\\ g\left( { - 1} \right) \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 1 - m = 0\\ 1 + 2 + m \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m = 1\\ m \ne - 3 \end{array} \right. \Leftrightarrow m = 1.\\ TH2:\,\,pt\,\,g\left( x \right) = {x^2} - 2x + m = 0\,\,co\,\,\,hai\,\,nghiem\,\,pb\,\,\,trong\,\,do\,\,co\,\,\,1\,\,nghiem\,\,\,x = - 1\\ \Leftrightarrow \left\{ \begin{array}{l} \Delta ' > 0\\ g\left( { - 1} \right) = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 1 - m > 0\\ 3 + m = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m < 1\\ m = - 3 \end{array} \right. \Leftrightarrow m = - 3\\ Vay\,\,\,m = 1;\,\,m = - 3\,\,thoa\,\,man. \end{array}$